If X is a Poisson variate such that 4P(X=2)=3P(X=1) , then the mean o…

The correct answer is Option (1) → 1.5

For a Poisson distribution with mean λ,

    \( \displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \)

So

    \( \displaystyle 4\cdot\frac{e^{-\lambda}\lambda^2}{2!}=3\cdot\frac{e^{-\lambda}\lambda}{1!} \)

Cancel \( e^{-\lambda}\lambda \) (both are non-zero):

    \( 2\lambda=3 \)

    \( \lambda=1.5 \)

So the mean of the distribution is 1.5.

Shortcut: Use the ratio of successive Poisson probabilities, \( \frac{P(X=2)}{P(X=1)}=\frac{\lambda}{2} \). Here the ratio is \( \frac{3}{4} \), so \( \lambda=\frac{3}{2}=1.5 \).